Question

NaOH, H2O, . O 2 0 CH,CHOH Hgc"CHE Acetone molas mass: 58 g/mol density: 0.79g/ml Ilmiting reagent original acetone carbons Benzaldehyde molar mass: 106 g/mol density: 1.04 g/mL an excess will be used Dibenzalacetone molar mass: 234 g/mol melting point: somewhere between 80 and 120°C

Answer 1

Given,

Molar mass of acetone = 58 g/mol

Moles of acetone = 0.0125 mol

Density of acetone = 0.79

We know that,

Molar mass of acetone × moles of acetone = Quantity of acetone Required

Therefore,

Quantity = 58 × 0.0125 = 0.725 g

Mass = 0.725 g

Density = Mass ÷ Volume

Therefore,  

Volume = Mass ÷ Density

Volume = 0.725 ÷ 0.79 = 0.917mL

Therefore 0.917 mL of volume is required to produce 0.0125 mol acetone.